Arguments of Roots of Complex Numbers

[I like number]

Homework Statement

i) Solve the equation z^3 = \mathbf{i}.
(ii) Hence find the possible values for the argument of a complex number w which is such that w^3 = \mathbf{i}(w*)^3.

I'm stuck on part ii.

Homework Equations

The Attempt at a Solution

The answer to the equation in part i: e^(1/6 pi i) , e^(5/6pi i ), e^(-1/2pi i )


In the back of the book it has 6 answers, I can only get three of them. If I draw it out on the argand diagram the 3 answers I am missing would be the arguments if you carried the line on through the axis. For example I have arguments 5/12pi , pi/12, -pi/4 and am missing 3/4pi, -11/12pi and -7/12pi.

My method was to cube root both sides:
e^theta i = Z e^ -theta i
Where Z is one of the solutions to part i, by changing Z I can get 3 solutions but am missing the other three.

After seeing in the book there were six solutions I did manage to get them by letting i = e^pi/2 i and setting it equal to e^6theta i, then I just kept adding 2pi to the argument of i until I got all 6 arguments and ended up back at the beginning. However if I didn't have the answers I wouldn't have known to look for the extra three, and this method didn't involve part i) of the question (which I think it is meant to because of the 'hence').

I'd be interested to see how you all approach the problem, and an explanation of what is actually happening.


Thanks

 

[tiny-tim]

Welcome to PF!

Hi I like number! Welcome to PF!

(have a pi: π and try using the X² tag just above the Reply box )

Hint: write it (w/w*)³ = i

 

[I like number]

Hi I like number! Welcome to PF!

(have a pi: π and try using the X² tag just above the Reply box )

Hint: write it (w/w*)³ = i

Thanks for the reply.
Doing that I end up with

e^2itheta=3rdrt i

From part i) of the question I have all 3 values for the 3rdrt i , so i set e^2itheta= to each of them one at a time , then find theta but this only gives 3 values.
If I was to plot the values I get on an argand diagram mine would be in the first and fourth quadrant, and I'm missing the 3 values from the second and third quadrants.

I hope I'm making sense, and thanks once again for the help.

I like number

 

[tiny-tim]

e^2itheta=3rdrt i

From part i) of the question I have all 3 values for the 3rdrt i …

ok, so you have three values for e^2iθ, which is (e^iθ)²,

so e^iθ is the square-root of that, which has two values, ± of each other.

 

[I like number]

ok, so you have three values for e^2iθ, which is (e^iθ)²,

so e^iθ is the square-root of that, which has two values, ± of each other.

*doh*!

e^iθ= ± e^π/12i
And from this I get the value I had which was π/12 if I take the positive root, if I take the negative root then the argument will be π12 + π = 13π/12 which gives the argument -11π/12 which was one of the missing ones :D If i do this for the others I'll get them :)
Perfect, thank you tiny-tim!

I like number